Key Concepts and Formulas

• Equation of a Circle (General Form): $x^2 + y^2 + 2gx + 2fy + c = 0$, where $(-g, -f)$ is the center and $r = \sqrt{g^2 + f^2 - c}$ is the radius.
• Equation of a Circle (Diameter Form): If $(x_1, y_1)$ and $(x_2, y_2)$ are endpoints of a diameter, then $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
• Slope of a Line: Given two points $(x_1, y_1)$ and $(x_2, y_2)$, the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Perpendicular Lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if $m_1 \cdot m_2 = -1$.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Identifying the Diameter Using Slopes

We are given four points $A(4,6)$, $B(-1,5)$, $C(0,0)$, and $D(k, 3k)$ that lie on a circle. We want to find if any pair of points forms a diameter. If two points form a diameter, then the angle subtended by these points at any other point on the circle will be $90^\circ$. This implies that the lines connecting the third point to the endpoints of the diameter are perpendicular.

Let's examine the slopes of lines formed by connecting point $B$ to $A$ and $C$:

• Slope of line $BA$: $m_{BA} = \frac{6 - 5}{4 - (-1)} = \frac{1}{5}$
• Slope of line $BC$: $m_{BC} = \frac{5 - 0}{-1 - 0} = -5$

Since $m_{BA} \cdot m_{BC} = \frac{1}{5} \cdot (-5) = -1$, lines $BA$ and $BC$ are perpendicular, which means $\angle ABC = 90^\circ$. Therefore, $AC$ is a diameter of the circle.

Step 2: Finding the Equation of the Circle

Since $A(4, 6)$ and $C(0, 0)$ are the endpoints of a diameter, we can use the diameter form of the equation of a circle: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ Substituting the coordinates of $A(4,6)$ and $C(0,0)$: $(x - 4)(x - 0) + (y - 6)(y - 0) = 0$ $x(x - 4) + y(y - 6) = 0$ $x^2 - 4x + y^2 - 6y = 0$ This is the equation of the circle.

Step 3: Calculating the Radius of the Circle

The center of the circle can be found by completing the square or by taking the midpoint of the diameter endpoints $A(4,6)$ and $C(0,0)$. The midpoint is $\left(\frac{4+0}{2}, \frac{6+0}{2}\right) = (2,3)$. Thus, the center of the circle is $(2,3)$.

The radius $r$ is the distance between the center $(2,3)$ and any point on the circle, such as $C(0,0)$. Using the distance formula: $r = \sqrt{(2 - 0)^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$ Therefore, $r^2 = 13$.

Step 4: Finding the Value of k

The point $D(k, 3k)$ also lies on the circle. Therefore, its coordinates must satisfy the equation of the circle: $x^2 - 4x + y^2 - 6y = 0$ Substitute $x = k$ and $y = 3k$: $k^2 - 4k + (3k)^2 - 6(3k) = 0$ $k^2 - 4k + 9k^2 - 18k = 0$ $10k^2 - 22k = 0$ $2k(5k - 11) = 0$ This gives us two possible solutions for $k$: $k = 0$ or $5k - 11 = 0 \Rightarrow k = \frac{11}{5}$. Since the problem states that the four points are distinct, $k$ cannot be 0 (otherwise $D$ would coincide with $C$). Thus, $k = \frac{11}{5}$.

Step 5: Calculating 10k + r^2

We have $k = \frac{11}{5}$ and $r^2 = 13$. Therefore: $10k + r^2 = 10\left(\frac{11}{5}\right) + 13 = 2(11) + 13 = 22 + 13 = 35$

Common Mistakes & Tips

• Assuming any three points define a circle: While true, choosing the right set of three points (or identifying a diameter) simplifies the calculations significantly.
• Forgetting the "distinct points" condition: Always check if the solution obtained satisfies all conditions given in the problem. In this case, we had to discard $k=0$ because it made point D identical to point C.
• Incorrectly applying the diameter form: Ensure you have correctly identified the endpoints of the diameter before plugging them into the formula.

Summary

We used the property that an angle subtended by a diameter at any point on the circle is a right angle to identify $AC$ as the diameter. Then, we used the diameter form to find the equation of the circle. Knowing a point on the circle, $D(k, 3k)$, allowed us to find the value of $k$, and finally, we calculated $10k + r^2$.

The final answer is \boxed{35}, which corresponds to option (C).